Problem: Divide the following complex numbers. $ \dfrac{3-4i}{3-4i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${3+4i}$ $ \dfrac{3-4i}{3-4i} = \dfrac{3-4i}{3-4i} \cdot \dfrac{{3+4i}}{{3+4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(3-4i) \cdot (3+4i)} {(3-4i) \cdot (3+4i)} = \dfrac{(3-4i) \cdot (3+4i)} {3^2 - (-4i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(3-4i) \cdot (3+4i)} {(3)^2 - (-4i)^2} = $ $ \dfrac{(3-4i) \cdot (3+4i)} {9 + 16} = $ $ \dfrac{(3-4i) \cdot (3+4i)} {25} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({3-4i}) \cdot ({3+4i})} {25} = $ $ \dfrac{{3} \cdot {3} + {-4} \cdot {3 i} + {3} \cdot {4 i} + {-4} \cdot {4 i^2}} {25} $ Evaluate each product of two numbers. $ \dfrac{9 - 12i + 12i - 16 i^2} {25} $ Finally, simplify the fraction. $ \dfrac{9 - 12i + 12i + 16} {25} = \dfrac{25 + 0i} {25} = 1 $